Question

For cach of the following balanced reactions, calculate how many moles of each product would be produced by complete conversion of 0.50 mole of the reactant indicated in boldface. Indicate clearly the mole ratio used for the conversion. a. \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\) b. \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) c. \(2 \mathrm{Al}(s)+6 \mathrm{HCl}(a q) \rightarrow 2 \mathrm{AlCl}_{3}(a q)+3 \mathrm{H}_{2}(g)\) d. \(\mathbf{C}_{3} \mathbf{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

a. \(0.50\,\text{moles}\,\mathrm{H}_2\mathrm{O}\) and \(0.25\,\text{moles}\,\mathrm{O}_2\) b. \(0.50\,\text{moles}\,\mathrm{KCl}\) and \(0.75\,\text{moles}\,\mathrm{O}_2\) c. \(0.50\,\text{moles}\,\mathrm{AlCl}_3\) and \(0.75\,\text{moles}\, \mathrm{H}_2\) d. \(1.5\,\text{moles}\,\mathrm{CO}_2\) and \(2\,\text{moles}\,\mathrm{H}_2\mathrm{O}\)

Step-by-step solution

Find the mole ratio

The mole ratio is obtained from the balanced reaction. Here, we have the ratios \(2 \mathrm{H}_{2}O_{2} : 2\mathrm{H}_2\mathrm{O} : \mathrm{O}_2\).

Convert moles of reactant to moles of products

We have 0.50 moles of \(\mathrm{H}_2 \mathrm{O}_2\), and we want to find the amount of moles of \(\mathrm{H}_2\mathrm{O}\) and \(\mathrm{O}_2\) produced using the mole ratios. For H2O: \(\frac{0.50 \,\text{moles}\, \mathrm{H}_2 \mathrm{O}_2 \cdot 2 \,\text{moles}\, \mathrm{H}_2\mathrm{O}}{2 \,\text{moles}\, \mathrm{H}_2 \mathrm{O}_2} = 0.50 \,\text{moles}\, \mathrm{H}_2\mathrm{O}\) For O2: \(\frac{0.50 \,\text{moles}\, \mathrm{H}_2 \mathrm{O}_2 \cdot 1 \,\text{moles}\, \mathrm{O}_2}{2 \,\text{moles}\, \mathrm{H}_2 \mathrm{O}_2} = 0.25 \,\text{moles}\, \mathrm{O}_2\) b. For the reaction \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\)

Find the mole ratio

The mole ratio is obtained from the balanced reaction. Here, we have the ratios \(2 \mathrm{KClO}_3 : 2\mathrm{KCl} : 3\mathrm{O}_2\).

Convert moles of reactant to moles of products

We have 0.50 moles of \(\mathrm{KClO}_3\), and we want to find the amount of moles of \(\mathrm{KCl}\) and \(\mathrm{O}_2\) produced using the mole ratios. For KCl: \(\frac{0.50 \,\text{moles}\, \mathrm{KClO}_3 \cdot 2 \,\text{moles}\, \mathrm{KCl}}{2 \,\text{moles}\, \mathrm{KClO}_3} = 0.50\,\text{moles}\, \mathrm{KCl}\) For O2: \(\frac{0.50 \,\text{moles}\, \mathrm{KClO}_3 \cdot 3 \,\text{moles}\, \mathrm{O}_2}{2 \,\text{moles}\, \mathrm{KClO}_3} = 0.75\,\text{moles}\, \mathrm{O}_2\) c. For the reaction \(2\mathrm{Al}(s)+6\mathrm{HCl}(aq)\rightarrow2\mathrm{AlCl}_{3}(aq)+3\mathrm{H}_{2}(g)\)

Find the mole ratio

The mole ratio is obtained from the balanced reaction. Here, we have the ratios \(2 \mathrm{Al} : 6\mathrm{HCl} : 2\mathrm{AlCl}_3 : 3\mathrm{H}_2\).

Convert moles of reactant to moles of products

We have 0.50 moles of \(\mathrm{Al}\), and we want to find the amount of moles of \(\mathrm{AlCl}_3\) and \(\mathrm{H}_2\) produced using the mole ratios. For AlCl3: \(\frac{0.50\,\text{moles}\,\mathrm{Al}\cdot2\,\text{moles}\,\mathrm{AlCl}_3}{2\,\text{moles}\,\mathrm{Al}}=0.50\,\text{moles}\,\mathrm{AlCl}_3\) For H2: \(\frac{0.50\,\text{moles}\,\mathrm{Al}\cdot3\,\text{moles}\,\mathrm{H}_2}{2\,\text{moles}\,\mathrm{Al}}=0.75\,\text{moles}\, \mathrm{H}_2\) d. For the reaction \(\mathrm{C}_3\mathrm{H}_8(g)+5\mathrm{O}_2(g)\rightarrow3\mathrm{CO}_2(g)+4\mathrm{H}_2\mathrm{O}(g)\)

Find the mole ratio

The mole ratio is obtained from the balanced reaction. Here, we have the ratios \(\mathrm{C}_3\mathrm{H}_8 : 5\mathrm{O}_2 : 3\mathrm{CO}_2 : 4\mathrm{H}_2\mathrm{O}\).

Convert moles of reactant to moles of products

We have 0.50 moles of \(\mathrm{C}_3\mathrm{H}_8\), and we want to find the amount of moles of \(\mathrm{CO}_2\) and \(\mathrm{H}_2\mathrm{O}\) produced using the mole ratios. For CO2: \(\frac{0.50\,\text{moles}\,\mathrm{C}_3\mathrm{H}_8\cdot3\,\text{moles}\,\mathrm{CO}_2}{1\,\text{moles}\,\mathrm{C}_3\mathrm{H}_8}=1.5\,\text{moles}\,\mathrm{CO}_2\) For H2O: \(\frac{0.50\,\text{moles}\,\mathrm{C}_3\mathrm{H}_8\cdot4\,\text{moles}\,\mathrm{H}_2\mathrm{O}}{1\,\text{moles}\,\mathrm{C}_3\mathrm{H}_8}=2\,\text{moles}\,\mathrm{H}_2\mathrm{O}\)

Key concepts

Chemical Reactions

Chemical reactions are the heart of chemistry, involving the transformation of reactants into products. In a chemical reaction, substances called reactants are transformed into different substances known as products. This transformation often involves a rearrangement of atoms, where bonds are broken and new bonds are formed. Every chemical reaction is unique, yet can be categorized into types like synthesis, decomposition, single replacement, and combustion. Each type follows distinct patterns and involves specific reactants and products. Understanding the type of reaction can often guide you in predicting the products. For instance, in a combustion reaction, a hydrocarbon like propane ( C_3H_8 ) burns in oxygen to form carbon dioxide ( CO_2 ) and water ( H_2O ). This transformation is not just about recognizing substances on the left (reactants) and right (products) of the equation but understanding the underlying changes happening at the molecular level. Mastery of chemical reactions is essential for predicting outcomes in experiments and industrial processes.

Mole Ratio

Mole ratio is a central concept in stoichiometry, providing the proportional relationship between reactants and products in a chemical reaction. It is derived from the balanced chemical equation, which gives the precise ratio of moles of one substance reacting with moles of another. By using the mole ratio, chemists can calculate the amounts of products formed from a given amount of reactant, or vice versa.
The balanced chemical equation serves as a mole map, guiding you in conversions. For example, if the balanced equation indicates that 2 moles of hydrogen peroxide ( H_2O_2 ) decompose to produce 2 moles of water ( H_2O ) and 1 mole of oxygen ( O_2 ), the mole ratio for water to oxygen is 2:1. This ratio is crucial when working with chemical quantities, as it ensures reactions are scaled appropriately, whether in a laboratory setting or large-scale production. Remember, the mole ratio directly connects the macroscopic world of measured substances to the microscopic world of chemical reactions.

Balancing Equations

Balancing chemical equations ensures the law of conservation of mass is obeyed, meaning the number of each type of atom on the reactant side must equal that on the product side. A balanced equation offers the correct stoichiometric coefficients, which are essential for calculating mole ratios. For instance, consider the equation of the decomposition of potassium chlorate: \(2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2\). Here, the balanced equation shows the disappearance of 2 moles of potassium chlorate to form 2 moles of potassium chloride and 3 moles of oxygen gas. Without a balanced equation, you cannot correctly compute mole ratios or predict the amount of products formed.
Balancing is often an iterative process, where you adjust coefficients to achieve balance. A handy tip is to balance one type of atom at a time, starting from the most complex molecule. This method ensures accuracy and ease when dealing with more complicated reactions. Mastering the skill of balancing equations is crucial for pursuing chemistry beyond the basics, as it lays the foundation for understanding all subsequent calculations and predictions in chemical contexts.